Question

abc=9000.(a,b)(b,c)(c,a) are pairs of co prime numbers.find a+b+c=?

Answer 1

Given equation is abc=9000.

Where pairs (a,b)(b,c)(c,a) are pairs of co prime numbers.

to find values of a,b,c first we need to find prime factorization of 9000. Which is given by:

9000 = 2*2*2*3*3*5*5*5

now we have to use those factors to form three numbers so that their product is 9000 keeping in mind that those three factors are coprime.

Two integers a and b are said to be relatively prime, mutually prime, or coprime (also spelled co-prime) if the only positive integer that divides both of them is 1.

So possible three such integers are (2*2*2), (3*3) , (5*5*5)

or 8, 9, 125

Hence a=8, b=9, c=125

Now we have to find value of a+b+c=8+9+125 = 8+9+125 =142

Hence final answer is a+b+c=142

Answer 2

Two numbers are said to be co-prime if the only common positive factor of the two numbers is 1.

Acc to the question, (abc)=9000 and (a,b)(b,c)(c,a) are pairs of co-prime numbers.

The factors of 9000 = 2³ * 3² * 5³ = 8 * 9 * 125

Therefore, a = 8; b = 9; c = 125.

Thus, a + b + c = 8 + 9 + 125 = 142. [Ans]